Midterm Exam Solutions - Computational Finance | ECON 424, Exams of Economics

Download Midterm Exam Solutions - Computational Finance | ECON 424 and more Exams Economics in PDF only on Docsity! University of Washington Fall 2007 Department of Economics Eric Zivot Economics 424 Midterm Exam Solutions This is a closed book and closed note exam. However, you are allowed one page of notes (double-sided). Answer all questions and write all answers in a blue book or on separate sheets of paper. Time limit is 1 hours and 50 minutes. Total points = 110. I. Return Calculations (25 pts, 5 points each) 1. Consider a one year investment in two Northwest stocks: Amazon and Costco. Suppose you buy Amazon and Costco at the end of September 2006 at , 1 , 1$32.12, $49.19A t C tP P− −= = and then sell at the end of September 2007 for , ,$93.15, $61.37A t C tP P= = . (Note: these are actual closing prices taken from Yahoo! The data for Amazon is not a mistake.) > pa.1 = 32.12 > pa.2 = 93.15 > pc.1 = 49.19 > pc.2 = 61.37 a. What are the simple annual returns for the two stocks? > ra = (pa.2 - pa.1)/pa.1 > rc = (pc.2 - pc.1)/pc.1 > ra [1] 1.9 > rc [1] 0.2476 b. What are the continuously compounded annual returns for the two stocks? > log(1 + ra) [1] 1.065 > log(1 + rc) [1] 0.2212 c. Costco paid the following per share cash dividends between September 2006 and September 2007: $0.13 in November, $0.13 in February, $0.145 in April, and $0.145 in July. What is the annual simple total return on Costco? What is the annual dividend yield? > rc.total = (pc.2 + 0.13 + 0.13 + 0.145 + 0.145 - pc.1)/pc.1 > div.y = (0.13 + 0.13 + 0.145 + 0.145)/pc.1 > rc.total [1] 0.2588 > div.y [1] 0.01118 > # total return = cap gain + div yeild rc + div.y [1] 0.2588 d. The annual inflation rate between September 2006 and September 2007 was about 3%. Using this information, determine the simple and continuously compounded real annual returns on Amazon and Costco. Note: for Costco, do not include the dividend adjustments. > inflat = 0.03 > # simple real returns ra.real = (1 + ra)/(1 + inflat) - 1 > rc.real = (1 + rc)/(1 + inflat) - 1 > ra.real [1] 1.816 > rc.real [1] 0.2113 > # cc real returns log(1 + ra.real) [1] 1.035 > log(1 + rc.real) [1] 0.1917 e. At the end of September, 2006, suppose you have $100,000 to invest in Amazon and Costco over the next year. Suppose you sell short $60,000 in Costco and use the proceeds to buy $160,000 in Amazon. Using the results from part a, compute the annual simple return on the portfolio. Assume that both stocks do not pay a dividend. > xc = -60000/100000 > xa = 160000/100000 > xa [1] 1.6 > xc [1] -0.6 > rp = xa * ra + xc * rc > rp [1] 2.892 > VaR.05 = (exp(q.05) - 1) * w0 > VaR.05 [1] -0.5619 > VaR.01 = (exp(q.01) - 1) * w0 > VaR.01 [1] -0.8641 III. Time Series Concepts (15 points) 1. Let { }tY represent a stochastic process. Under what conditions is { }tY covariance stationary? (5 points) 2 [ ] var( ) cov( , ) (depends on and not ) t t t t j j E Y Y Y Y j t μ σ γ− = = = 2. Consider the random walk model 1 0 0 1 2 , constant. ~ (0, ) t t t t j j t Y Y Y Y iid N ε ε ε σ − = = + = + =∑ Is { }tY a covariance stationary stochastic process? Why or why not? (5 points) No. The random walk process is not stationary. The variance of the random walk process depends on time: 2 2 1 1 var( ) var( ) t t t j j j Y tε σ σ = = = = =∑ ∑ 3. The figure below shows annual observations on the dividend yield of the S&P 500 index over the period 1871 through 2000 along with the sample ACF. (5 points) Annual dividend price ratio on S&P 500 1870 1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 0. 02 0. 06 0. 10 Lag AC F 0 5 10 15 20 -0 .2 0. 2 0. 4 0. 6 0. 8 1. 0 Series : dpratio a) Assume the dividend yield is covariance stationary. Based on the shape of the sample autocorrelation function, would an MA(1) process or an AR(1) process best describe the data? Briefly justify your answer. The SACF decays toward zero and does not cut off at lag 1. Therefore, it looks more like an AR(1) process than an MA(1) process. VI. Constant Expected Return Model (40 points, 5 points each) Consider the constant expected return model 2, ~ (0, ) cov( , ) , ( , ) it i it it i it jt ij it jt ij r iid N r r corr r r μ ε ε σ σ ρ = + = = for the monthly continuously compounded returns on the Vanguard extended market index (vexmx) and the Vanguard long-term bond index (vbltx) (subset of class project data) over the period September 2002 through September 2007. For this period there are T=60 monthly observations. The data are shown in the graph below. -0 .0 5 0. 00 0. 05 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 2002 2003 2004 2005 2006 2007 vexmx -0 .0 5 0. 00 0. 05 vbltx a) Do the monthly continuously compounded return data look like they come from the CER model? Why or why not? The CER model postulates that cc returns are (covariance stationary) iid normal random variables with constant means, variances and covariances (correlations). The above two return series look a bit like computer simulations from the CER model. The returns appear to fluctuate randomly about a constant mean. However, the volatility appears to be slightly higher before 2004 than after 2004 suggesting that the variances of the returns are not constant through time. b) What are the estimators (formulas used to compute estimates) for 2, , and i i ijμ σ σ ? 2 2 1 1 1 1 1ˆ ˆ ˆ, ( ) 1 1ˆ ˆ ˆ( )( ) T T i t i tt t T ij it i jt jt r r T T r r T μ σ μ σ μ μ = = = = = − − = − − ∑ ∑ ∑ f) Test the following hypotheses using a 5% significance level: 0 1: 0 vs. : 0vexmx vexmxH Hμ μ= ≠ ; 0 1: 0 vs. : 0vbltx vbltxH Hμ μ= ≠ 0 , 1 ,: 0 vs. : 0vexmx vbltx vexmx vbltxH Hρ ρ= ≠ . Here, we can test hypotheses in two ways: (1) use t- statistics; (2) use 95% confidence intervals. Using t- statistics we have > t.stat.mu0 = muhat.vals/se.muhat > t.stat.rho0 = rhohat.vals/se.rho > abs(t.stat.mu0) vexmx vbltx 3.339 1.281 > abs(t.stat.rho0) vexmx,vbltx 0.9736 Since T=60, our rule-of-thumb decision rule is: reject the null that the true value is zero at the 5% level if the absolute value of the t-statistic is greater than 2. We reject the null only for vexmx. Using 95% confidence intervals, our decision rule is: reject the null that the true value is zero at the 5% level if zero is not in the 95% confidence interval. From the previous question, we see that zero is not in the 95% confidence interval only for vexmx. g) The figures below gives some graphical diagnostics of the return distributions for vexmx and vbltx. Also, estimated values of the skewness and excess kurtosis for vexmx and vbltx are vexmx vbltx skewness -0.1111852 -0.9670961 excess kurtosis -0.4891066 2.4082389 -0.05 0.0 0.05 0.10 0 2 4 6 8 10 VEXMX monthly returns rvexmx -0 .0 5 0. 0 0. 05 monthly return de ns ity e st im at e -0.05 0.0 0.05 0.10 0 2 4 6 8 10 Quantiles of Standard Normal V E X M X re tu rn s -2 -1 0 1 2 -0 .0 5 0. 0 0. 05 -0.10 -0.05 0.0 0.05 0 5 10 15 vbltx monthly returns rvbltx -0 .0 8 -0 .0 2 0. 04 monthly return de ns ity e st im at e -0.10 -0.05 0.0 0.05 0 5 10 15 Quantiles of Standard Normal vb ltx re tu rn s -2 -1 0 1 2 -0 .0 8 -0 .0 2 0. 04 Based on this information, do you think the monthly cc returns on vexmx and vbltx are normally distributed? Briefly justify your answer. For vexmx, the graphical diagnostics are consistent with a normal distribution: the histogram is bell shaped, the box plot is symmetric with no outliers and the qq-plot against the normal distribution is linear. Additionally, the sample skewness and excess kurtosis values are close to zero. If we compute the JB statistics we get > JB = (nobs * (vexmx.skew^2 + 0.25 * vexmx.ekurt^2))/6 > JB [1] 0.7217 Since JB < 6, we do not reject the null hypothesis that the returns on vexmx are normally distributed. The story is different for vbltx. Here the graphical diagnostics indicate some departures from the normal distribution. This histogram is negatively skewed (long left tail), the boxplot shows one moderate outlier, and the qq-plot deviates from linearity in the lower tail. The sample skewness is fairly negative and the excess kurtosis is quite large. The JB statistic is > JB = (nobs * (vbltx.skew^2 + 0.25 * vbltx.ekurt^2))/6 > JB [1] 23.85 which is greater than 6 so we reject the null hypothesis at the 5% level that the returns are normally distributed. i) Below are the sample autocorrelation functions for vexmx and vbltx. Using the information in these graphs, would you say that the CER model assumption that returns are uncorrelated over time is appropriate? Briefly justify your answer.